Question

A manufacturer ships parts in lots of 1000 and makes a profit of $50 per lot sold. The purchaser, however, subjects the product to a sampling plan as follows: 10 parts are selected at random with replacement. If none of these parts is defective, the lot is purchased; if one part is defective,the lot is purchased but the manufacturer returns $10 to the buyer; if two or more parts are found to be defective, the entire lot is returned at a net loss of $25 to the manufacturer. What is the manufacturer’s expected profit if 10% of the parts are defective?

Answer 1

Given that,

10% of the parts are defective out of 1000

probability of getting a defective part (p) = 0.1
probability of getting a non defective part (1-p) = 1 - 0.1 =0.9

we know that, probability of binomial distribution is P(X=x) = ⁿC_x (p)^x (1-p)^n-x

Now a sample of 10 parts (with replacement) is taken by the manufacturer, there are 3 possible outcomes

1) A: none of 10 part is defective Profit(A) = $50
P(A) = ¹⁰C₀ (p)⁰ (1-p)¹⁰ = 1 * 1 * 0.9¹⁰ = 0.3486

2)B: exactly 1 part is defective Profit(B) = $40
P(B) = ¹⁰C₁ (p)¹ (1-p)⁹ = 10 * 0.1 * 0.9⁹ = 0.3874

3)C: 2 or more parts are defective Profit(C) = $-25
P(B) = 1 - P(A) - P(B) = 1 - 0.3486 - 0.3874 = 0.2639

Expected profit = P(A)*Profit(A) + P(B)*Profit(B) + P(C)*Profit(C) = 0.3486*50 + 0.3874*40 + 0.2639*(-25) = $26.3285

The manufacturer’s expected profit if 10% of the parts are defective is $26.3285